$f(t) = 2t+4$ $h(n) = -2n^{2}-4n-5(f(n))$ $g(n) = 7n^{3}-2n^{2}-4n+2(f(n))$ $ g(f(-2)) = {?} $
Solution: First, let's solve for the value of the inner function, $f(-2)$ . Then we'll know what to plug into the outer function. $f(-2) = (2)(-2)+4$ $f(-2) = 0$ Now we know that $f(-2) = 0$ . Let's solve for $g(f(-2))$ , which is $g(0)$ $g(0) = 7(0^{3})-2(0^{2})+(-4)(0)+2(f(0))$ To solve for the value of $g$ , we need to solve for the value of $f(0)$ $f(0) = (2)(0)+4$ $f(0) = 4$ That means $g(0) = 7(0^{3})-2(0^{2})+(-4)(0)+(2)(4)$ $g(0) = 8$